1064 Complete Binary Search Tree (30分)

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805407749357568

题目

1064 Complete Binary Search Tree (30分)

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

A Complete Binary Tree (CBT) is a tree that is completely filled, with the possible exception of the bottom level, which is filled from left to right.

Now given a sequence of distinct non-negative integer keys, a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤1000). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.

Output Specificationx:

For each test case, print in one line the level order traversal sequence of the corresponding complete binary search tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input:

10
1 2 3 4 5 6 7 8 9 0

 

Sample Output:

6 3 8 1 5 7 9 0 2 4

解题思路

题意：给定一个数列，要求输出该序列数构成的完全二叉搜索树。

完全二叉搜索树（完全数与二叉排序数综合，CBST）：在满足完全二叉树的基础上，非叶节点n的左子树元素均小于n的元素值，而右子树大于n的元素值。而完全二叉树的层序遍历具有的性质为，第i个元素输出下标为i(i从1计数)。

性质：对于下标x，2*x与2*x+1为x的左右孩子节点下标（注意下标 1<= index <=n）

思路：1、对于完全二叉搜索树，该树的中序遍历结果即为这些元素的非递减数列（及该数列的排序结果）。

           2、现在题目要求，给定数列，来求该树的层序遍历，我们可以采用逆推的方式，以前我们是已知CBST来进行中序遍历，那么我们是否可以考虑任然采用中序遍历，只是对于CBST re[x]来说，我们虽然未知，但是我们知道该点元素值为num[ct]，由此我们使用逆推法，根据中序遍历结果以及CSBT的性质，来得到我们re[x]中的原有元素值。

           3、我们将给定的数列进行排序，得到排序结果数组（CBST）：num

           4、此时num[1]输出的数，即为CBST中序遍历的第一个数，依次类推得到所有的结果。

算法建模：

         ct:该元素输出顺序

         num：元素数列非递减排序

        re:CBST，re[i] 第i个元素

        x:下标

       中序遍历 dfs:(x<=n)

          dfs(2*x)  //先左子树

          re[x] = ct++;//num  中的x 输出的顺序为ct

          dfs(2*x+1) //遍历右子树

 

解题代码

#include <bits/stdc++.h>

using namespace std;
//完全搜索二叉树
//
const int maxsize = 1010;
int re[maxsize];
int num[maxsize];
int n;
int ct = 1;
void dfs(int x){
    if(x>n) return ;
    dfs(2*x);
    /*
        中序遍历：printf("%d",num[x])
        Xi元素输出的顺序：ct
    */
    re[x] = ct++;//第一个输出的元素序号
    dfs(2*x+1);
}
void init(){
    cin>>n;
    for(int i = 1 ;i <=n;i++){
        cin>>num[i];
    }
    sort(num+1,num+n+1);
    int sum = n;
    dfs(1);
    for(int i = 1 ; i <=n;i++){
        cout<<num[re[i]];
        if(i!=n) cout<<" ";
    }
}
int main()
{
    init();
    return 0;
}